The sum to infinity of the given series frac{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given series is $\frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3} - \frac{1}{4n^4} + \dots$ This can be rewritten as $\frac{1}{n} - \frac{(1/n)^2

Vedclass · Accepted Answer

$\log_e\left(\frac{n+1}{n}\right)$

Vedclass · Answer

(A) The given series is $\frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3} - \frac{1}{4n^4} + \dots$ This can be rewritten as $\frac{1}{n} - \frac{(1/n)^2}{2} + \frac{(1/n)^3}{3} - \frac{(1/n)^4}{4} + \dots$ We know the logarithmic series expansion $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$ for $|x| Substituting $x = \frac{1}{n}$,we get $\log_e(1 + \frac{1}{n}) = \frac{1}{n} - \frac{(1/n)^2}{2} + \frac{(1/n)^3}{3} - \dots$ Therefore,the sum is $\log_e\left(\frac{n+1}{n}\right)$.

The sum to infinity of the given series $\frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3} - \frac{1}{4n^4} + \dots$ is

Similar Questions

The expression $\log_{e} 2 + \log_{e} \left( 1 + \frac{1}{2} \right) + \log_{e} \left( 1 + \frac{1}{3} \right) + \dots + \log_{e} \left( 1 + \frac{1}{n - 1} \right)$ is equal to

The value of the infinite series $\log _4 2 - \log _8 2 + \log _{16} 2 - \dots \infty$ is:

For $|x| < 1$,the coefficient of $x^3$ in the expansion of $\log(1+x+x^2)$ in ascending powers of $x$ is (in $/3$)

$1 + \frac{(\log_e n)^2}{2!} + \frac{(\log_e n)^4}{4!} + \dots = $

The sum of $\frac{1}{2} + \frac{1}{3} \cdot \frac{1}{2^3} + \frac{1}{5} \cdot \frac{1}{2^5} + \dots \infty$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module